If `x=3i-6j-k, y=i+4j-3k" and "z=3i-4j-12k`, then the magnitude of the projection of `x times y` on z is

Options

A. 12
B. 15
C. 14
D. 13

Answer

To find the magnitude of the projection of \( \mathbf{x} \times \mathbf{y} \) on \( \mathbf{z} \), we will follow these steps:

### Step 1: Compute the cross product \( \mathbf{x} \times \mathbf{y} \)

Given:
\[ \mathbf{x} = 3\mathbf{i} - 6\mathbf{j} - \mathbf{k} \]
\[ \mathbf{y} = \mathbf{i} + 4\mathbf{j} - 3\mathbf{k} \]

The cross product \( \mathbf{x} \times \mathbf{y} \) can be calculated using the determinant of a matrix:

\[ \mathbf{x} \times \mathbf{y} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -6 & -1 \\ 1 & 4 & -3 \end{vmatrix} \]

Calculating the determinant:
\[ = \mathbf{i} \begin{vmatrix} -6 & -1 \\ 4 & -3 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 3 & -1 \\ 1 & -3 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 3 & -6 \\ 1 & 4 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} -6 & -1 \\ 4 & -3 \end{vmatrix} = (-6)(-3) - (-1)(4) = 18 + 4 = 22 \)
2. \( \begin{vmatrix} 3 & -1 \\ 1 & -3 \end{vmatrix} = (3)(-3) - (-1)(1) = -9 + 1 = -8 \)
3. \( \begin{vmatrix} 3 & -6 \\ 1 & 4 \end{vmatrix} = (3)(4) - (-6)(1) = 12 + 6 = 18 \)

Putting it all together:
\[ \mathbf{x} \times \mathbf{y} = 22\mathbf{i} + 8\mathbf{j} + 18\mathbf{k} \]

### Step 2: Compute the dot product \( (\mathbf{x} \times \mathbf{y}) \cdot \mathbf{z} \)

Given:
\[ \mathbf{z} = 3\mathbf{i} - 4\mathbf{j} - 12\mathbf{k} \]

Now compute the dot product:
\[ (\mathbf{x} \times \mathbf{y}) \cdot \mathbf{z} = (22\mathbf{i} + 8\mathbf{j} + 18\mathbf{k}) \cdot (3\mathbf{i} - 4\mathbf{j} - 12\mathbf{k}) \]
\[ = 22 \cdot 3 + 8 \cdot (-4) + 18 \cdot (-12) \]
\[ = 66 - 32 - 216 \]
\[ = 66 - 32 - 216 = -182 \]

### Step 3: Compute the magnitude of \( \mathbf{z} \)

\[ |\mathbf{z}| = \sqrt{3^2 + (-4)^2 + (-12)^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13 \]

### Step 4: Compute the magnitude of the projection

The magnitude of the projection of \( \mathbf{x} \times \mathbf{y} \) on \( \mathbf{z} \) is given by:
\[ \text{Magnitude of projection} = \frac{|(\mathbf{x} \times \mathbf{y}) \cdot \mathbf{z}|}{|\mathbf{z}|} \]
\[ = \frac{|-182|}{13} = \frac{182}{13} \approx 14 \]

### Final Answer
The magnitude of the projection of \( \mathbf{x} \times \mathbf{y} \) on \( \mathbf{z} \) is \( \frac{182}{13} \) or approximately \( 14 \).

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